Consider: note that (check the initial values and the recurrence relation)

Now the RHS of your identity equals

First note that, if a vector satisfies then ( indeed, at most we can have n-1 2s ) (1)

So it is formed by combining a vector of entries, and a complementary one, like this: ( can be 0, or )

Now choose of the first entries to be 2s, the rest of the first n entries are going to be defined as 1s. We can do this in ways. And we have summed so far. So we require . but this can be done in exactly: ways. So if, among the first n entries, we have 2s, we can achieve this in ways.

If we sum all the possible number of 2s among the first n entries ( disjoint cases) we'll have: